Dear friend, The expression 

((√1 + y′ ² )/dy')

cannot be solved. Only equations can be solved. Softwares normally solve an expression by equating the expression to zero. If that is the way you want, yes maple can solve it.

((√1 + y′ ² )/dy') = 0 -> √1 + y′ ²  = 0  -> 1 + y′ ²  = 0 -> y′ ²= -1 ->  y′ =  i -> dy = i dx -> y = i x²/2

Cheers. 

Ramakrishnan V

Dear Sir, The plot starts for x value zero correspondingly, the y value (integral is infinity).

There is a distinct difference from 0.0001 and 0.00001 and correspondingly the y values are finite and infinite as can be seen from the maple evluated values corresponding to x values. An appropriate x values clearly indicate this. Change the x limits to 0.0001 to 0.00015 and you will see the plotted values clearly.

Your third execution of procedure gives the integral values for a given x and y value, whereas the plot is for an entire range of x values from zero to 0.00015. Corresponding plot  shows y value from infinity (asymptote can be drawn) to close to zero.

Hope the explanation is clear and satisfactory.

Ramakrishnan V.

Download Int_Question_answer.mw

Please permit me to explain a little I know about the Newton's method of solving equations. The solutions or Roots of an equation are zero values of the expression. (Ex: 2x = 3 is an equation (bring all terms to left hand side) , then 2x-3 is the expression and the root or zero value or solution of the equation is x = - 3/2).

If you plot the expression, it will either cut the (real number line) x axis or not. You will get chaotic in the later case because there is no solution.

In case of an expression that has a zero value (i.e. solution), if you plot the expression, then the slope at any point tend to point towards the (x axis) real line closer to the zero value (in real line, f(x) = 0 at that x value)

A closer look at the curve (what ever function be it) will tell you that at any point (value of x you take as an initial value, f(x) is an opposite side (from zero point) and slope is opposite side by adjacent side. Remember that slope is nothing but tan =  df(x)/dx.Hence (f(x)/tan = ), Reciprocal of secant (i.e. adjacent length to reach zero value from current initial value) directly gives the interval.

The interval keeps reducing with subsequent calculations.

Here a is initial value, ab is adjacent length, b is the new x value closer to zero value and r is the zero value. 

Therefore b = a - [f(a)/f'(a)]       [Note secant line is f(a)/f'(a) = ab]

If your initial assumption for x0 (solution) is closer to r (zero value), it will be very simple and quick for manual calculations. For computer, it is immaterial and takes extra few milliseconds!!.

Hope I have not confused you.

Cheers. Ramakrishnan V

Mathematics does it automatically only  the way you wanted.

Download PowerOfIntegers.mw

In my opinion, it acts like a procedure statement

As you have stated,PDEtools[declare](y(t)) tells Maple that y is a function of t, and therefore y(t) is displayed as y and the derivative of y is displayed as y_t.

The procedure p(s) tells Maple that p is a function of s, and therefore p(s) is displayed as p and the derivative of p is displayed as p_s.

Hope I am clear

Cheers.

Ramakrishnan V

For evn values of n, straight away the result can be taken as the fixed value 1.

For odd values of n, nx^n,= nx by nature since nx = +1 or -1 only.

For evn values of n, nx^n = 1 by nature since -1^odd is -1 and +1^odd is +1 and nx = +1 and -1 only

The problem as I understand is simple and as follows.

In MATLAB the solution is

n = [1 2 3 4];
nx = [1 -1];
z = [nx(1).^n;nx(2).^n];

z =

     1     1     1     1  (for nx = 1)
    -1     1    -1     1  (for nx = -1)

The first row gives all answers as 1 ^ any number is 1.

The second row gives -1 ^odd number is -1 and -1 ^ even number is even 

Hope I have not mistaken the problem.

Much easier commands must be available in Maple for this matlab solution.

Cheers.

Ramakrishnan V

@wsningrat 

Though I am a novice in this area, I just thought giving my idea would sometimes lead you to a solution.

In matlab, when you have all the solutions, both wanted (negative) and unwanted (positive), there is a quick way to discard unwanted elements and retain the remaining elements.

The commands used are logical (indexing) operation and conditional operations.

First select all the elements that are negative and index them.

Example: Let z = {(x,y)}be [[1, +0.5], [3, +0.8], [4, -0.8],[5, +0.4],[6, -0.3]]

Indexing is here z(:,2)<0; i.e. z(3) and z(5). Now idx = (3,5) are indexes.

(: referes to all rows; 2 refers to 2nd column in all rows (+ and -ve values here)

The solution is z(~idx) . Nonindexed z values are the required solution.

(Here ~ refers to not condition and  idx refers to indexed values)

There must be a way to do this in MAPLE as well.

Cheers.

Ramakrishnan V

Dear Sir,

Hope attached doc helps you in some way for solution.

printf format as given below is used to produce the print out.
printf("\n             Report on  %5.55s    %d", l, m);
 

Download printValues_whileLoop.mw

printValues_whileLoop.mw

Cheers. Ramakrishnan V

Please find attached document following the order of commands.

y = ax+b is the format in first line

x,y data are given in second line. (The data you give as (xn, xn+1){ which exercise you can do yourself from the given x values, arranging it in ascending order first)

Command for curve fitting in third line

Print the equation command in fourth line

Plot the graph of the line in fifth command

You can also plot your given values in the plot and compare!

Thanks to Dr.Robert J Lopez who taught me this and Maplesoft.

Cheers.

Ramakrishnan V
 

Download Fix_plot_stLineData.mw

There is no mistake in the for loop;

Only in the procedure the statement " uses RonanRoutines" is a mistake.

Further your programme works in the first instance only; second time it does not work for this error statement.

There is no match for this "RonanRoutines" word in the help section also.

nops(soll) remains 4 and the for loop executes 4 times as well.

Hope I may be of some use.

Proc_for_Pedal_not_working_(1).mw
 

Download Proc_for_Pedal_not_working_(1).mw

Cheers.

Ramakrishnan V

restart is very important

The maple commands recommended will work in a fresh document, otherwise  (or unassign r and theta, may be, but I have not used) restart command is essential.

I enclose the doc which works fine, but does not work second time if restart is removed (even) from the first line. It is theredore very important to use restart in the beginning of any document.

I have used the same first commands group three times. Second time wihout restart and third time with restart. Second time it gives error, but third one, it answers. 

It is theredore very important to use restart even at places (as required)  in the middle of any document.

Note: To get the the animation,  click on the plot to get the animation menu and then select press play icon.

Download polarplotexample.mw

Cheers.

Ramakrishnan V

Dear user,

You have got the answer in the first line itself. The only oversight is the correct bracket curly one for the list and not square bracket as shown in your question. Commands and doc attached below for reference. I must mention Dr.Robert Lopez taught me this (types of brackets for specific uses).

To get the answer, assign a:=; a[1] gives you [1,2].
Cheers.

Ramakrishnan V

Download IntersectExample.mw

IntersectExample.mw

Dear user,

First, define the polynomial (curve) format as y = an x^n+....+constant.

Next, give a list as d = [[2,2],[12,6]....]

Then use the package command here, c = CurveFitting[LeastSquare](d,x,curve = curveformat)

Thats all. You get the resulting equation.

Hope the attached document commands help you understand the commands.

curveFit_plotPoints.mw
 

Download curveFit_plotPoints.mw

I learnt this from Prof.Dr.Robert Lopez. Thanks Professor.

Cheers.

Ramakrishnan V

Four equations and four unknowns gives you the answers as given below which includes in third solution,  the simple answer you suggested..

Four equations and Five equations gives you more answers including the particular solution when a > 0, d = 0. Hence when a > 0, d is not {R-0}.

Ramakrishnan V

Download mapleprime10102018.mw

mapleprime10102018.mw

Hope my doc here may help you solve your problem.

Thanks.

Ramakrishnan V
 

Download arrowplots.mw